Edexcel International A Level Chemistry: Chirality (YCH11)

This guide teaches Topic 15A: Chirality of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers outcomes 15.1 to 15.5. It opens Topic 15: Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality, and is listed under Unit 4: Rates, Equilibria and Further Organic Chemistry, one of the International A2 (IA2) units. It draws on the halogenoalkane mechanisms from Unit 2 and the rate equations of Topic 11.

Useful pages: the shorter chirality revision notes; the chirality practice questions; a printable checklist; the course hub; and a free diagnostic to find your gaps. For the other kind of stereoisomerism in this course, geometric (E/Z) isomerism, see the alkenes study guide.

Topic 15A at a glance

Outcome	What the outcome asks of you
15.1	Know that optical isomerism comes from chirality in molecules with a single chiral centre
15.2	Link optical isomerism to chiral centres (asymmetric carbon atoms); describe enantiomers as object and non-superimposable mirror images; draw 3D diagrams of them
15.3	Know that optical activity is the ability of a single optical isomer to rotate the plane of polarisation of plane-polarised monochromatic light
15.4	Know what a racemic mixture is
15.5	Use optical activity data on reactants and products as evidence for SN1 and SN2 mechanisms and for addition to carbonyl compounds

15.1 and 15.2: Chirality and chiral centres

An object is chiral if it cannot be superimposed on its own mirror image, like a left and a right hand.

In organic molecules, chirality usually comes from a chiral centre (an asymmetric carbon atom): a carbon bonded to four different atoms or groups. The four tetrahedral bonds allow two arrangements that are mirror images. These two arrangements are optical isomers, also called enantiomers.

Optical isomerism is a type of stereoisomerism: the isomers share one structural formula, but their atoms are arranged differently in space.

How to find a chiral centre

1. Ignore every CH₃ carbon, every CH₂ carbon and every carbon in a C=C or C=O. None of these can have four different groups.
2. For each remaining carbon, list the four groups attached. Write each group out in full (for example C₂H₅, not just “a carbon”).
3. Compare the groups as whole groups, not just the first atom. CH₃ and C₂H₅ both attach through carbon but are different groups.
4. If all four are different, the carbon is a chiral centre. Mark it with an asterisk (*).

Worked example 1: one chiral centre or none?

Decide whether each molecule has a chiral centre.

Molecule	Carbon checked	Four groups	Chiral centre?
3-methylheptane, CH₃CH₂CH(CH₃)CH₂CH₂CH₂CH₃	C3	H, CH₃, C₂H₅, C₄H₉	Yes
4-methylheptane, CH₃CH₂CH₂CH(CH₃)CH₂CH₂CH₃	C4	H, CH₃, C₃H₇, C₃H₇	No: two propyl groups
butane-1,3-diol, HOCH₂CH₂CH(OH)CH₃	C3	H, OH, CH₃, CH₂CH₂OH	Yes
3-hydroxybutanal, CH₃CH(OH)CH₂CHO	C3	H, OH, CH₃, CH₂CHO	Yes

The trap is 4-methylheptane: the chain on each side of C4 is a propyl group, so there is no chiral centre.

Worked example 2: more than one chiral centre

Outcome 15.2 says chiral centre(s), so you should be able to spot more than one. Take 2,3-dichloropentane, CH₃CHClCHClCH₂CH₃.

• C2 carries H, Cl, CH₃ and CHClCH₂CH₃. Four different groups, so C2 is chiral.
• C3 carries H, Cl, C₂H₅ and CHClCH₃. Four different groups, so C3 is chiral.

The molecule has two chiral centres: CH₃CHClCHClCH₂CH₃. The “rest of the molecule” counts as one whole group.

15.2: Drawing optical isomers in 3D

You must be able to draw both enantiomers so that the 3D shape is clear. Use the standard conventions:

• a plain line shows a bond lying in the plane of the paper;
• a solid wedge is a bond coming towards you;
• a hashed (dashed) wedge is a bond going away from you.

A tetrahedral carbon is usually drawn with two plain bonds, one solid wedge and one hashed wedge. Then draw the mirror image beside it, reflected across a vertical mirror line: whatever is on the left in the first drawing appears on the right in the second.

Worked example 3: the enantiomers of 2-chlorobutanoic acid

2-chlorobutanoic acid is CH₃CH₂CHClCOOH. Its C2 carries H, Cl, C₂H₅ and COOH, so it is a chiral centre and the molecule has two optical isomers.

  Enantiomer A                mirror               Enantiomer B
                                 :
           COOH                  :                  HOOC
          /                      :                      \
  C2H5 -- C ===> Cl  (wedge)     :     (wedge)  Cl <=== C -- C2H5
          |                      :                      |
          H  (hashed)            :            (hashed)  H

In A, C₂H₅ and COOH lie in the plane, Cl points towards you and H points away. B is A’s reflection. If you rotate B to line up H and Cl with A, the ethyl and acid groups end up swapped, so A and B cannot be superimposed: they are enantiomers.

Checks for a good 3D drawing:

• the central carbon shows a tetrahedral arrangement (two in-plane bonds, one wedge, one hashed bond);
• each group is joined by its correct atom (the C of COOH, not the O, is bonded to the chiral carbon);
• the second drawing is a true mirror image, not the same molecule redrawn.

15.3: Optical activity

Two enantiomers have the same melting temperature, boiling temperature and density. The difference shows up with plane-polarised light: light that a polarising filter has restricted to vibrations in one plane. The light used is monochromatic (a single wavelength), for example from a sodium lamp, because the angle of rotation depends on the wavelength.

Optical activity is the ability of a single optical isomer to rotate the plane of polarisation of plane-polarised monochromatic light.

How the measurement works (a polarimeter):

1. Monochromatic light passes through a fixed polarising filter, giving plane-polarised light.
2. The light passes through a tube holding a solution of the sample.
3. A second filter (the analyser) is turned to find the new plane. The angle turned, compared with pure solvent, is the angle of rotation.

Under the same conditions (same concentration, path length, solvent, temperature and wavelength), each enantiomer turns the plane through an equal angle, but in the opposite direction to the other. One is labelled (+), turning the plane clockwise as you look towards the light source; the other is (−), turning it anticlockwise. You cannot tell which enantiomer is (+) from a 3D drawing alone. The sign has to be measured.

15.4: Racemic mixtures

A racemic mixture (a racemate) is a mixture containing equal amounts of the two enantiomers of a compound. It is not optically active: each (+) molecule’s rotation is cancelled by a (−) molecule’s equal and opposite rotation, so the net rotation is zero.

Do not confuse the two reasons for “no rotation”:

• a racemic mixture contains chiral molecules, but their rotations cancel;
• an achiral compound (no chiral centre) has no optical isomers at all, so it never rotates the plane.

The reading is zero in both cases; only the structure tells them apart.

15.5: Optical activity as evidence for mechanisms

This is where Topic 15A links to Topic 11 kinetics and to carbonyl chemistry. The logic is always the same: what does the mechanism do to the shape around the chiral carbon, and is the product a single enantiomer or a racemic mixture?

SN1: a planar carbocation gives a racemic mixture

In SN1 hydrolysis of a tertiary halogenoalkane, the rate-determining step is the breaking of the C–halogen bond to form a carbocation. That carbocation is trigonal planar. The nucleophile can attack from either face of the plane with equal probability.

Worked example 4: hydrolysis of 3-chloro-3-methylheptane

A single enantiomer of 3-chloro-3-methylheptane, CH₃CH₂C(CH₃)Cl(CH₂)₃CH₃, is warmed with aqueous sodium hydroxide. The product, 3-methylheptan-3-ol, shows no optical activity. Explain this.

1. The reactant’s C3 carries CH₃, C₂H₅, C₄H₉ and Cl, so it is chiral. The single enantiomer is optically active.
2. The product’s C3 carries CH₃, C₂H₅, C₄H₉ and OH, so the product is chiral too. A lack of rotation therefore cannot come from the product being achiral.
3. The C–Cl bond breaks first, giving a planar tertiary carbocation.
4. OH⁻ attacks either face equally, forming the two enantiomers of 3-methylheptan-3-ol in equal amounts.
5. The product is a racemic mixture, so its rotations cancel. This supports an SN1 mechanism.

SN2: backside attack gives a single, inverted enantiomer

In SN2, the nucleophile attacks the carbon from the side opposite the halogen as the C–halogen bond breaks, in one step. The other three groups flip over, like an umbrella turning inside out. This is inversion of configuration. Only one enantiomer of the product forms.

Worked example 5: hydrolysis of 2-iodohexane

One enantiomer of 2-iodohexane, CH₃CHI(CH₂)₃CH₃, reacts with aqueous hydroxide ions. The hexan-2-ol formed is optically active. Which mechanism does this support?

1. C2 in the reactant carries H, I, CH₃ and C₄H₉: chiral. C2 in the product carries H, OH, CH₃ and C₄H₉: chiral.
2. An optically active product means it is (mainly) a single enantiomer, not a racemic mixture.
3. A single enantiomer forms when OH⁻ attacks from the side opposite the iodine, inverting the arrangement. This supports SN2.

Secondary halogenoalkanes such as 2-iodohexane can react by either route, which is why optical data are useful: they show which route operated. Do not predict the product’s sign of rotation from the reactant’s sign; it must be measured.

Nucleophilic addition to carbonyl compounds

In aldehydes and ketones the carbonyl carbon is trigonal planar. In the addition of HCN (with KCN present), the cyanide ion attacks the carbonyl carbon. It can approach from either side of the plane with equal probability.

Worked example 6: HCN with 5-methylhexan-2-one

5-methylhexan-2-one, CH₃COCH₂CH₂CH(CH₃)₂, reacts with HCN in the presence of KCN. The product is 2-hydroxy-2,5-dimethylhexanenitrile, CH₃C(OH)(CN)CH₂CH₂CH(CH₃)₂. Predict and explain whether the product is optically active.

1. The new C2 carries CH₃, OH, CN and CH₂CH₂CH(CH₃)₂: four different groups, so the product is chiral.
2. The ketone is planar at the C=O carbon, and CN⁻ attacks from above or below the plane with equal probability.
3. Each enantiomer is made in the same amount, giving a racemic mixture.
4. The product is not optically active. This fits a mechanism with attack on a planar carbonyl group.

Compare 2,6-dimethylheptan-4-one, (CH₃)₂CHCH₂COCH₂CH(CH₃)₂. Its HCN product has two identical 2-methylpropyl groups on the new carbon, so it has no chiral centre at all. Optical data could not give evidence about the mechanism for that ketone.

Summary of the evidence

Mechanism	Shape at the reacting carbon	Product from a single enantiomer (or planar reactant)	Rotation
SN1	Planar carbocation intermediate	Racemic mixture	None
SN2	Five-group transition state; inversion	Single enantiomer	Yes
Nucleophilic addition to C=O	Planar C=O attacked from either side	Racemic mixture (if product is chiral)	None

These results fit the Topic 11 rate equations: rate = k[halogenoalkane] for SN1 and rate = k[halogenoalkane][OH⁻] for SN2.

Common errors

• Counting a CH₂ carbon or a C=O carbon as a chiral centre.
• Judging groups by their first atom alone, so CH₃ and C₂H₅ get treated as “the same”.
• Drawing two pictures that are the same molecule rotated, rather than mirror images.
• Writing “the product is racemic, so it has no chiral centre”. A racemic mixture is made of chiral molecules.
• Leaving out “monochromatic” or “plane-polarised” in a definition of optical activity.
• Writing “SN1 because the product is inactive” with no planar intermediate or attack from both sides.

Where next

Use the revision notes, then the practice questions. Naming and structural isomerism are in the introductory organic chemistry guide.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, published by Pearson Education Limited. Topic 15A: Chirality, outcomes 15.1 to 15.5.